First we start with the homogeneous system. The coefficient matrix is\begin{equation} A={\left[\begin{array}{ccc}1 & 1 \\-1 & 1 \end{array} \right ]},\end{equation}The eigenvalues are\begin{equation} \lambda_1=1+i\end{equation}\begin{equation} \lambda_2=1-i\end{equation}The eigenvector corresponding to $\lambda_1$ is \begin{equation}\xi^(1)=(1,i)^{T}\end{equation}Then we can have the solution to the homogeneous system\begin{equation}(x,y)^T=c_1\cdot (e^t\cos(t), -e^t\sin(t))^T+c_2\cdot(e^t\sin(t), e^t\cos(t))^T\end{equation}We can get a fundamental matrix from here\begin{equation} \Psi={\left[\begin{array}{ccc}e^t\cos(t) & -e^t\sin(t) \\e^t\sin(t) & e^t\cos(t)\end{array} \right ]},\end{equation}Note that we have \begin{equation} g(t)=(\frac{e^t}{\cos(t)},\frac{e^t}{\sin(t)})^T\end{equation}And suppose\begin{equation} \Psi \mu^\prime=g(t)\end{equation}We shoud have for the nonhomogeneous system\begin{equation}(x,y)^T=\Psi \mu\end{equation}This is the method of Variation of Parameters. By plugging in and integration, we have\begin{equation} \mu_1(t)=c_3\end{equation}\begin{equation} \mu_2(t)=-\ln|cot(2t)+csc(2t)|+c_4\end{equation}Finally we get the required solution according to (6), (9), (10) and (11)